Scientific Method Worksheet for 4th - 5th Grade | Lesson Planet - Free Printable

It looks like you've uploaded a text-based image containing a mathematical problem, but the actual image isn't visible to me. However, based on your description and the text provided, I can help solve the problem if you clarify or describe it in detail.

From the text you've shared, it appears to be a mathematical logic or set theory problem involving:

- A set of numbers (possibly integers) where certain conditions are applied.
- A function $ f(n) $ defined on positive integers, possibly related to digit manipulation or number properties.
- The goal seems to be analyzing the behavior of such a function or proving something about its outputs.

Let’s try to reconstruct what might be the problem from the text snippet:

> "From the laws of logic, Aristotle has discovered that the world is based on a logical structure... Let $ f(n) $ be a function defined on positive integers such that $ f(n) = n + 1 $ if $ n $ is odd, and $ f(n) = n/2 $ if $ n $ is even."

This resembles the Collatz function, which is defined as:
$$
f(n) =
\begin{cases}
n/2 & \text{if } n \text{ is even} \\
3n+1 & \text{if } n \text{ is odd}
\end{cases}
$$

But here, it's slightly different: $ f(n) = n+1 $ if odd, $ f(n) = n/2 $ if even.

Wait — let's look again at the text:

> "Let $ f(n) $ be a function defined on positive integers such that $ f(n) = n + 1 $ if $ n $ is odd, and $ f(n) = n/2 $ if $ n $ is even."

So this is not the standard Collatz function. It's a custom-defined function.

Now, suppose the task is to analyze this function: perhaps find the behavior of iterating $ f $, or determine whether every starting number eventually reaches 1, or find cycles, etc.

Let’s assume the problem is:

---

Problem:

Define a function $ f: \mathbb{Z}^+ \to \mathbb{Z}^+ $ by:
$$
f(n) =
\begin{cases}
n+1 & \text{if } n \text{ is odd} \\
n/2 & \text{if } n \text{ is even}
\end{cases}
$$
Study the behavior of repeated applications of $ f $. Does every positive integer eventually reach 1? Are there cycles?

---

Step-by-step Solution:

We'll compute $ f(n) $ iteratively for small values of $ n $ to see what happens.

#### Try $ n = 1 $:
- $ f(1) = 1+1 = 2 $ (odd → add 1)
- $ f(2) = 2/2 = 1 $ (even → divide by 2)
- $ f(1) = 2 $
- $ f(2) = 1 $

So we get a cycle: $ 1 \to 2 \to 1 \to 2 \to \dots $

So cycle exists: $ \{1, 2\} $

#### Try $ n = 3 $:
- $ f(3) = 4 $ (odd → +1)
- $ f(4) = 2 $ (even → /2)
- $ f(2) = 1 $
- $ f(1) = 2 $ → enters the cycle

So: $ 3 \to 4 \to 2 \to 1 \to 2 \to \dots $

Eventually reaches the cycle.

#### Try $ n = 5 $:
- $ f(5) = 6 $
- $ f(6) = 3 $
- $ f(3) = 4 $
- $ f(4) = 2 $
- $ f(2) = 1 $
- Then cycles.

So: $ 5 \to 6 \to 3 \to 4 \to 2 \to 1 \to 2 \to \dots $

#### Try $ n = 7 $:
- $ f(7) = 8 $
- $ f(8) = 4 $
- $ f(4) = 2 $
- $ f(2) = 1 $
- Cycle reached.

#### Try $ n = 9 $:
- $ f(9) = 10 $
- $ f(10) = 5 $
- $ f(5) = 6 $
- $ f(6) = 3 $
- $ f(3) = 4 $
- $ f(4) = 2 $
- $ f(2) = 1 $
- Cycle.

Seems like all numbers eventually reach the $ \{1,2\} $ cycle.

But wait — is this always true?

Let’s check $ n = 15 $:
- $ f(15) = 16 $
- $ f(16) = 8 $
- $ f(8) = 4 $
- $ f(4) = 2 $
- $ f(2) = 1 $
- Cycle.

What about an odd number that keeps growing?

Suppose $ n $ is odd and large: $ f(n) = n+1 $, which is even. Then $ f(n+1) = (n+1)/2 $. So we go from odd $ n $ to $ n+1 $, then down to $ (n+1)/2 $.

Compare $ (n+1)/2 $ with $ n $: For $ n > 1 $, $ (n+1)/2 < n $ when $ n > 1 $. So the value decreases after two steps.

For example:
- $ n = 15 $: $ 15 \to 16 \to 8 $ → much smaller than 15
- $ n = 101 $: $ 101 \to 102 \to 51 $ → still less than 101

So the sequence tends to decrease over time.

Only exception: if $ n = 1 $, it goes to 2, then back to 1.

So no number escapes to infinity.

Also, since the values are decreasing eventually, and positive integers are bounded below by 1, the process must eventually enter a cycle.

We already found one cycle: $ 1 \leftrightarrow 2 $

Are there others?

Suppose there's another cycle.

Try to find fixed points or cycles.

#### Fixed point: $ f(n) = n $

Case 1: $ n $ odd → $ f(n) = n+1 = n $? → impossible.

Case 2: $ n $ even → $ f(n) = n/2 = n $ → $ n = 0 $, not positive.

So no fixed points.

#### 2-cycle: $ a \to b \to a $

Try:
- $ a $ odd → $ f(a) = a+1 = b $
- $ b $ even → $ f(b) = b/2 = a $

So: $ b = a+1 $, and $ b/2 = a $ → $ (a+1)/2 = a $ → $ a+1 = 2a $ → $ a = 1 $

Then $ b = 2 $

So only 2-cycle is $ 1 \to 2 \to 1 $

Any longer cycle?

Try 3-cycle: $ a \to b \to c \to a $

Assume $ a $ odd → $ b = a+1 $ (even)

$ b $ even → $ c = b/2 = (a+1)/2 $

Now $ c $ could be odd or even.

Then $ f(c) = a $

So $ f(c) = a $

If $ c $ odd → $ f(c) = c+1 = a $

So $ c+1 = a $ → $ (a+1)/2 + 1 = a $ → $ (a+1 + 2)/2 = a $ → $ (a+3)/2 = a $ → $ a+3 = 2a $ → $ a = 3 $

Then $ c = (3+1)/2 = 2 $, even → contradiction (we assumed $ c $ odd)

So invalid.

If $ c $ even → $ f(c) = c/2 = a $

So $ c/2 = a $ → $ (a+1)/2 / 2 = a $? Wait:

$ c = (a+1)/2 $, so $ f(c) = c/2 = (a+1)/4 $

Set equal to $ a $: $ (a+1)/4 = a $ → $ a+1 = 4a $ → $ 3a = 1 $ → $ a = 1/3 $, not integer.

So no 3-cycle.

Similarly, higher cycles would require more complex chains, but due to the halving step, values drop quickly.

Thus, the only cycle is $ \{1,2\} $

And from all tested values, sequences eventually reach this cycle.

Can we prove that every positive integer eventually reaches the cycle $ \{1,2\} $?

Yes, using induction or monotonicity.

Note:
- If $ n $ is odd: $ f(n) = n+1 $ → even
- Then $ f(n+1) = (n+1)/2 $

So from any odd $ n $, next step is $ (n+1)/2 $

Compare $ (n+1)/2 $ with $ n $: for $ n \geq 2 $, $ (n+1)/2 < n $

For $ n = 1 $: $ f(1) = 2 $, then $ f(2) = 1 $, stays in cycle.

So all odd $ n \geq 3 $ lead to smaller numbers.

Even $ n $: $ f(n) = n/2 $, so strictly smaller unless $ n = 2 $

So the sequence either decreases or enters the cycle.

Hence, no infinite increasing chain possible, and since values are positive integers, the process must eventually reach the cycle $ \{1,2\} $.

---

Conclusion:

The function $ f(n) $ defined by:
$$
f(n) =
\begin{cases}
n+1 & \text{if } n \text{ is odd} \\
n/2 & \text{if } n \text{ is even}
\end{cases}
$$
has the property that for every positive integer $ n $, repeated application of $ f $ eventually enters the cycle $ 1 \to 2 \to 1 \to \dots $.

There are no other cycles, and no number escapes to infinity.

---

Final Answer:

✔ Every positive integer eventually reaches the cycle $ \boxed{\{1, 2\}} $ under repeated application of $ f $.

If your original problem was different (e.g., a specific value to compute, or a proof), please clarify, and I’ll adjust accordingly!